Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{r^2 - 2r - 24}{r - 4} \times \dfrac{-6r + 24}{9r + 36} $
Explanation: First factor the quadratic. $z = \dfrac{(r + 4)(r - 6)}{r - 4} \times \dfrac{-6r + 24}{9r + 36} $ Then factor out any other terms. $z = \dfrac{(r + 4)(r - 6)}{r - 4} \times \dfrac{-6(r - 4)}{9(r + 4)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (r + 4)(r - 6) \times -6(r - 4) } { (r - 4) \times 9(r + 4) } $ $z = \dfrac{ -6(r + 4)(r - 6)(r - 4)}{ 9(r - 4)(r + 4)} $ Notice that $(r - 4)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -6\cancel{(r + 4)}(r - 6)(r - 4)}{ 9(r - 4)\cancel{(r + 4)}} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $z = \dfrac{ -6\cancel{(r + 4)}(r - 6)\cancel{(r - 4)}}{ 9\cancel{(r - 4)}\cancel{(r + 4)}} $ We are dividing by $r - 4$ , so $r - 4 \neq 0$ Therefore, $r \neq 4$ $z = \dfrac{-6(r - 6)}{9} $ $z = \dfrac{-2(r - 6)}{3} ; \space r \neq -4 ; \space r \neq 4 $